Set up Taylor expansion #\bb\color(red)\text(formula)# for #sqrt(x)# around #a=2#?
Check my work please?
#f'=(1/2)x^(-1/2)#
#f''=(1/2)(-1/2)x^(-3/2)#
#f'''=(1/2)(-1/2)(-3/2)x^(-5/2)#
#f^4=(1/2)(-1/2)(-3/2)(-5/2)x^(-7/2)#
#f^5=(1/2)(-1/2)(-3/2)(-5/2)(-7/2)x^(-9/2)#
#f'(2)=(1/2)*2^(-1/2)#
#f''(2)=(1/2)(-1/2)*2^(-3/2)#
#f''(2)'=(1/2)(-1/2)(-3/2)*2^(-5/2)#
#f^4(2)=(1/2)(-1/2)(-3/2)(-5/2)*2^(-7/2)#
#f^5(2)=(1/2)(-1/2)(-3/2)(-5/2)(-7/2)*2^(-9/2)#
Okay, I don't know where to go from here.
but I DO know that the exponents increase in this order...
#(-1/2)+(-1)^\beta# (not sure if #\beta=n# yet)
Check my work please?
Okay, I don't know where to go from here.
but I DO know that the exponents increase in this order...
1 Answer
Explanation:
For this function, finding the
We will not evaluate the derivatives at
If we look only at the derivatives and disregard
That is, we start with
The first derivative has no negative sign, the second derivative is negative, and this pattern continues for the following derivatives. Since we start at
The exponent on the
In the denominator, we
Now, for the numerator, we see we have the pattern
We can represent this with the double factorial,
Thus,
Now, we use the general form for a Taylor series centered at
to see that
Note that we did not include
We can't simplify the factorials any further.